Here's the probability (I think) that a particle in Brownian motion (w/ standard deviation $\sqrt{t}$) will exceed $m$ between times $t_1$ and $t_2$:

$$\frac1{2\sqrt{2\pi}}\int_{-\infty }^m \frac1{\sqrt{t_1}}e^{-\frac{x^2}{2t_1}}\left(1+\mathrm{erf}\left(\frac{m-x}{2\sqrt{t_2-t_1}}\right)\right)\mathrm{d}x$$

or, in Mathematica (slightly different form):

p[m_,t1_,t2_] := Integrate[ 
 CDF[NormalDistribution[x, Sqrt[2]*Sqrt[t2-t1]]][m], 
{x,-Infinity,m}, Assumptions -> { 
 t2 >= t1 >= 0, Element[m,Reals], Element[t2, Reals], Element[t1, Reals]} 

Mathematica can numerically integrate this for specific values of m, t1, and t2, but it's not superfast.

I now want to find the partial derivatives of p[] with respect to each of its variables. Ideally in closed-form (Mathematica can't find one), but a good approximation if not.

I've tried power series and a few other techniques, but I've found nothing good enough for a wide range of values for m, t1, and t2.

Solving this problem will calculate the "Greeks" for box options:


  • $\begingroup$ Perhaps, there is something wrong in your expression for that probability. I will post an answer soon. $\endgroup$ – Shai Covo Nov 18 '10 at 23:38
  • $\begingroup$ You do know the error function is built-in in Mathematica as Erf[]? $\endgroup$ – J. M. is a poor mathematician Nov 19 '10 at 0:22
  • $\begingroup$ Yes, I used Mathematica to generate the TeX. I'm not sure why I didn't use it in the Mathematica code, but it shouldn't affect Mathematica's ability to integrate. $\endgroup$ – barrycarter Nov 19 '10 at 17:42

In view of your previous post, the probability you are considering is actually $$ \bar P(m,t_1,t_2) = 1 - \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2 - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x},\;\; m \in {\bf R}. $$ This is the complement of the probability considered in your previous post. So, we may consider instead the partial derivatives of $$ P(m,t_1,t_2) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2 - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x}. $$ Now, erf(x) has derivative ${\rm erf}'(x)= \frac{2}{{\sqrt \pi }}e^{ - x^2 }$, $x > 0$. Then, the partial derivative with respect to $t_2$ is obtained as follows: $$ \frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial t_2 }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2 - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x}, $$ and explicitly, $$ \frac{{\partial}}{{\partial t_2 }}P(m,t_1 ,t_2 ) = \frac{{ - 1}}{{\sqrt {8 \pi ^2 t_1 (t_2 - t_1 )^3 } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2 - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg](m - x){\rm d}x}. $$ (The probability that the maximum is smaller than $m$ decreases as $t_2$ increases, hence the minus sign in this equation.) To find the partial derivative with respect to $m$ corresponds to taking the derivative of a convolution, according to the rule $(f * g)' = f' * g$. Specifically, this says that $$ \frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) = \frac{1}{{\sqrt {2\pi t_1 } }}\int_{ - \infty }^m {\frac{\partial }{{\partial m }}{\rm erf}\bigg(\frac{{m - x}}{{2\sqrt {t_2 - t_1 } }}\bigg)e^{ - x^2 /(2t_1 )} {\rm d}x}, $$ and explicitly, $$ \frac{{\partial}}{{\partial m }}P(m,t_1 ,t_2 ) = \frac{{1}}{{\sqrt {2 \pi ^2 t_1 (t_2 - t_1 ) } }}\int_{ - \infty }^m {\exp \bigg[ - \frac{{(m - x)^2 }}{{4(t_2 - t_1 )}} - \frac{{x^2 }}{{2t_1 }}\bigg]{\rm d}x}. $$ Maybe you can simplify further. The partial derivative with respect to $t_1$ is apparently a quite complicated expression.

  • 2
    $\begingroup$ The partial derivative with respect to $m$ is equivalent to $\frac{m^2}{\sqrt{2\pi(2t_2-t_1)}}\exp\left(-\frac{m^2}{2(2t_2-t_1)}\right)\left(1+\mathrm{erf}\left(m\sqrt{\frac{t_2-t_1}{t_1(2t_2-t_1)}}\right)\right)$ . The other one looks quite complicated. $\endgroup$ – J. M. is a poor mathematician Nov 19 '10 at 2:36
  • $\begingroup$ Elegant expression! Thank you. $\endgroup$ – Shai Covo Nov 19 '10 at 3:04
  • $\begingroup$ Thanks! I'd actually been looking at the CDF for m, so meant to say "less than or equal to m", but you are correct. I wonder if that's any way to coerce Mathematica into coming up with this answer. $\endgroup$ – barrycarter Nov 19 '10 at 17:55
  • $\begingroup$ Since I can scale my time unit, I'm wondering if I can set t1 to 1 without loss of generality, hmm... (or t2-t1 to one, and leaving t1 arbitrary) $\endgroup$ – barrycarter Nov 19 '10 at 20:14
  • $\begingroup$ OK, I think I'm confused by the @/@m derivative, since m is a limit of the integral. Would the fundamental theorem of calculus apply? $\endgroup$ – barrycarter Nov 19 '10 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.